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CBSE Class 9 Mathematics Question 9 of 9

Number Systems — Question 3

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Question 3

Simplify:

(i) (223)\left(2^{\smash{\frac{2}{3}}} \right). (215)\left(2^{\smash{\frac{1}{5}}} \right)

(ii) (133)7\Big(\dfrac{1}{3^3}\Big)^7

(iii) 11121114\dfrac{11^\frac{1}{2}}{11^\frac{1}{4}}

(iv) (712)\left(7^{\smash{\frac{1}{2}}} \right). (812)\left(8^{\smash{\frac{1}{2}}} \right)

Answer

(i) (223)\left(2^{\smash{\frac{2}{3}}} \right). (215)\left(2^{\smash{\frac{1}{5}}} \right)

= (223+15)\left(2^{\smash{\frac{2}{3} + \frac{1}{5}}} \right)

= ((2)10+315)\left((2)^{\smash{\frac{10+3}{15}}} \right)

= ((2)1315)\left((2)^{\smash{\frac{13}{15}}} \right)

Hence, (223)\left(2^{\smash{\frac{2}{3}}} \right). (215)\left(2^{\smash{\frac{1}{5}}} \right) = ((2)1315)\left((2)^{\smash{\frac{13}{15}}} \right)

(ii) (133)7\Big(\dfrac{1}{3^3}\Big)^7

=(133)7=(1733×7)=(1321)= \Big(\dfrac{1}{3^3}\Big)^7 \\[1em] = \Big(\dfrac{1^7}{3^{3 \times 7}}\Big) \\[1em] = \Big(\dfrac{1}{3^{21}}\Big)

Hence, (133)7=(1321)\Big(\dfrac{1}{3^3}\Big)^7 = \Big(\dfrac{1}{3^{21}}\Big)

(iii) 11121114\dfrac{11^\frac{1}{2}}{11^\frac{1}{4}}

= (1112−14)\left(11^{\smash{\frac{1}{2} - \frac{1}{4}}} \right)

= (112−14)\left(11^{\smash{\frac{2-1}{4}}} \right)

= (1114)\left(11^{\smash{\frac{1}{4}}} \right)

Hence, 11121114\dfrac{11^\frac{1}{2}}{11^\frac{1}{4}} = (1114)\left(11^{\smash{\frac{1}{4}}} \right)

(iv) (712)\left(7^{\smash{\frac{1}{2}}} \right). (812)\left(8^{\smash{\frac{1}{2}}} \right)

=(7×8)12[∵am×bm=(ab)m]=5612= (7 \times 8)^{\dfrac{1}{2}}\quad[\because \text{a}^m \times \text{b}^m = (\text{ab})^m] \\[1em] = 56^{\frac{1}{2}}

Hence, (712).(812)=5612\left(7^{\smash{\frac{1}{2}}} \right).\left(8^{\smash{\frac{1}{2}}} \right) = 56^{\frac{1}{2}}

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Mathematics | Number SystemsWeb Content • Interactive Notes

Number Systems — Interactive Study Guide

Master the real number system, irrational numbers, surds, rationalisation, and exponent laws.

The Number Hierarchy

Think of numbers as nested boxes: Natural numbers are inside Whole numbers, which are inside Integers, which are inside Rational numbers, which are inside Real numbers.

Key Insight: Between any two rational numbers, there are infinitely many irrational numbers, and vice versa. The number line is “dense” with both types!

Identifying Rational vs Irrational

NumberTypeReason
√4Rational√4 = 2 (perfect square)
√7Irrational7 is not a perfect square
0.333...RationalRecurring decimal = 1/3
0.10100100010...IrrationalNon-terminating, non-recurring
πIrrationalNon-terminating, non-recurring
22/7RationalIt is p/q form (just an approximation of π)

Rationalisation — Quick Method

To rationalise a denominator with surds, multiply top and bottom by the conjugate:

  • Conjugate of (a + √b) is (a − √b)
  • Conjugate of (√a − √b) is (√a + √b)

The denominator becomes rational because (a+b)(a−b) = a² − b².

Quick Self-Check

  1. Is √(16/9) rational or irrational? (Rational: = 4/3)
  2. Simplify: √50 + √18 (= 5√2 + 3√2 = 8√2)
  3. Rationalise: 1/(√3 + 1) (= (√3 − 1)/2)
  4. Find: 81/3 (= 2)

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